Lecture 20

Solution: Clearly $\mathrm{Ind}_{S_{4}}^{S_{6}}\left(\pi,{\mathcal{S}}^{(3,1)}\right)\cong\mathrm{Ind}_{S_{5}}^{S_{6}}\mathrm{Ind}_{S_{4}}^{S_{5}}\left(\pi,{\mathcal{S}}^{(3,1)}\right)$ as making a choice of coset representatives for $S_{4}\leq S_{5}$ and then for $S_{5}\leq S_{6}$ gives us a set of coset representatives for $S_{4}\leq S_{6}$ and induction is independent of coset representation, by Problem 81. Thus, by Theorem 20.4,

Solution: Recall that $V\cong\ker(T)\oplus{\mathrm{im}}(T)$ as vector spaces so we simply need to show that the $G$-action is preserved. Define

which is an isomorphism of vector spaces. Let $g\in G$, then

for all ${\bf v}\in V$. Thus $\phi$ is a $G$-homomorphism and hence a $G$-isomorphism.

Solution: First we note that $T_{i}^{\prime}$ is defined on the basis of ${\mathcal{M}}^{\lambda}$ so it is clearly linear. Let $\sigma\in S_{n-1}$ and $[t]\in{\mathcal{M}}^{\lambda}$. We note that $[t]$ and $[\sigma\cdot t]$ both have $n$ in the same row as all elements of $S_{n-1}$ leave $n$ invariant. Thus we have $\sigma\cdot T_{i}^{\prime}([t])=0=T_{i}^{\prime}([\sigma\circ t])$, when $n$ and $\mathsf{c}_{i}$ are in different rows and

otherwise. Thus $T_{i}^{\prime}$ is an $S_{n-1}$-homomorphism.

Note that $T_{1}$ is a restriction of a linear map, hence linear and thus, by induction, each $T_{i}$ is linear. Let $\sum_{[t]\in{\mathcal{M}}^{\lambda}}\alpha_{[t]}[t]\in{\mathcal{S}}^{\lambda}$, for $\alpha_{[t]}\in{\mathbb{C}}$. Then, as ${\mathcal{S}}^{\lambda}$ is a subrepresentation $\sigma\cdot\sum_{[t]\in{\mathcal{M}}^{\lambda}}\alpha_{[t]}[t]\in{\mathcal{S}}^{\lambda}$ and

for all $\sigma\in S_{n-1}$. Thus $T_{1}$ is a $S_{n-1}$-homomorphism. For the inductive step suppose $T_{i-1}$ is an $S_{n-1}$-homomorphism and let $\sum_{[t]{\mathrm{im}}{\mathcal{M}}^{\lambda}}\alpha_{[t]}[t]\in\ker T_{i-1}$, for $\alpha_{[t]}\in{\mathbb{C}}$. Then, as $\ker T_{i-1}$ is a subrepresentation $\sigma\cdot\sum_{[t]{\mathrm{im}}{\mathcal{M}}^{\lambda}}\alpha_{[t]}[t]\in{\mathcal{S}}^{\lambda}$ and

for all $\sigma\in S_{n-1}$. Thus, by induction, $T_{i}$ is an $S_{n-1}$-homomorphism for each $i$.

Solution: Let $w\in\mathrm{im}(T)$. We may therefore write $w=Tv$ for some $v\in V$. For any $g\in G$, we have

where we used that $T\in\operatorname{Hom}_{G}(V,W)$ for the second equality. The above computation shows that $\mathrm{im}(T)$ is $\rho(G)$- invariant, and thus $(\rho,W)$ is a subrepresentation of $(\rho,W)$.

Solution:

1. (a)
   ​

   Since $M^{d}=I$, we have $M^{d}-I=\prod_{j=0}^{d}(M-e^{2\pi ij/d}I)=0.$ The minimal polynomial of $M$ must therefore divide $\prod_{j=0}^{d}(M-e^{2\pi ij/d}I)$, and hence factors as a number of distinct linear factors $M-e^{2\pi ij/d}I$. This implies that $M$ is diagonalisable.

   Alternatively, write $M$ in Jordan normal form:

   $$M=B\operatorname{diag}\big{(}J_{k_{1}}(\lambda_{1}),J_{k_{2}}(\lambda_{2}),\ldots J_{k_{m}}(\lambda_{m})\big{)}B^{-1},$$

   where each $J_{k}(\lambda)$ is a $k\times k$ Jordan block:

   $$J_{k}(\lambda)=\left(\begin{matrix}\lambda&1&0&0&\cdots&0\\ 0&\lambda&1&0&\cdots&0\\ \vdots&&\ddots&\ddots&&\vdots\\ 0&0&0&0&\cdots\lambda&1\\ 0&0&0&0&\cdots&\lambda\end{matrix}\right)$$

   Thus, taking the $d$-th power, we have

   $$I=M^{d}=B\operatorname{diag}\big{(}J_{k_{1}}(\lambda_{1})^{d},J_{k_{2}}(\lambda_{2})^{d},\ldots J_{k_{m}}(\lambda_{m})^{d}\big{)}B^{-1}.$$

   We must therefore have

   $$J_{k_{i}}(\lambda_{i})^{d}=I_{k_{i}}$$

   for all $i=1,\ldots,m$. Writing $J_{k}(\lambda)=\lambda I_{k}+N_{k},$ where $N_{k}$ is the $k\times k$ matrix with ones on the superdiagonal, and all other entries zero, we have

   $$J_{k}(\lambda)^{d}=(\lambda I_{k}+N_{k})^{d}=\lambda^{d}I_{k}+\sum_{i=1}^{d}\left(\begin{matrix}n\\ i\end{matrix}\right)\lambda^{d-i}N_{k}^{i}.$$

   Note now that $N_{k}^{i}$ has ones on the $i$-th superdiagonal, and all other entries zero (and $N_{k}^{k}=0$). This means that $N_{k}$ cannot be written as a linear combination of the other $N_{k}^{i}$s; and so the only way for $J_{k}(\lambda)^{d}=I_{k}$ to hold is if $k=1$. If all Jordan blocks are $1\times 1$ matrices then $M$ is diagonalisable.

2. (b)
   ​

   then follows from (a) by choosing a basis ${\mathcal{B}}$ of $V$ and considering the matrix of $\pi(g)$ with respect to this basis. Since $\pi(g)$ is of finite order, so is its matrix w.r.t. any choice of basis (cf. Problem 19), hence the matrix of $\pi(g)$ is diagonalisable by part (a). Using the change of basis formula to change the basis ${\mathcal{B}}$ to that basis given by the diagonalisation results in the claimed eigenbasis of $\pi(g)$.

Proposition 15. Let $(\pi,V)$ be a finite-dimensional representation of a finite group $G$. For any irreducible representation $(\rho,W)$ of $G$ and any decomposition of $(\pi,V)$ into irreducible subrepresentations, the number of components in the decomposition that are isomorphic to $(\rho,W)$ is equal to $\operatorname{dim}\big{(}\operatorname{Hom}_{G}(V,W)\big{)}$.

Why does this complete the proof of Theorem 13?

Solution: By the first part of Theorem 13, given a representation $(\pi,V)$, we may write it as a direct sum of irreducible subrepresentations:

where each $(\pi,W_{i})$ is irreducible. Given an irreducible representation $(\rho,W)$, we then have

where Lemma 14 (i) was used for the second equality. Since the dimension of a direct sum is the sum of the dimensions, this gives

By Corollary 7, $\operatorname{dim}(\operatorname{Hom}_{G}(W_{i},W))=1$ if $(\pi,W_{i})\cong(\rho,W)$, and zero otherwise, hence

This proves the proposition. The last part of Theorem 13 follows since the right-hand side of the above equation is independent of the choice of decomposition of $(\pi,V)$ into irreducible subrepresentations.

The following problems give an alternative proof of Maschke’s Theorem (Theorem 10): Recall that an inner product $\langle\cdot,\cdot\rangle$ on a vector space $V$ is a map $V\times V\rightarrow{\mathbb{C}}$ such that

1. (i)
   ​

   $\langle v,w\rangle=\overline{\langle w,v\rangle}$ for all $v,w\in V$ (conjugate symmetry).

2. (ii)
   ​

   $\langle\alpha u+\beta v,w\rangle=\alpha\langle u,w\rangle+\beta\langle v,w\rangle$ for all $\alpha,\beta\in{\mathbb{C}}$, $u,v,w\in V$ (linearity in the first argument).

3. (iii)
   ​

   $\langle v,v\rangle>0$ if $v\neq 0$ (positive-definiteness).

Recall that the standard example on ${\mathbb{C}}^{n}$ is as follows:

Solution: Let $\{v_{i}\}_{i=1,\ldots,\operatorname{dim}V}$ be any basis of $V$. We define $\langle\cdot,\cdot\rangle:V\times V\rightarrow{\mathbb{C}}$ by

This is well-defined due to $\{v_{i}\}_{i=1,\ldots,\operatorname{dim}V}$ being a basis: any pair of vectors in $V$ can be written in a unique way as a linear combination of the $v_{i}$s.

It remains to verify that this is indeed an inner product:

1. (i)
   $$\left\langle\sum_{i=1}^{\operatorname{dim}V}a_{i}v_{i},\sum_{j=1}^{\operatorname{dim}V}b_{j}v_{j}\right\rangle=\sum_{i=1}^{n}a_{i}\overline{b_{i}}=\overline{\sum_{i=1}^{n}b_{i}\overline{a_{i}}}=\overline{\left\langle\sum_{j=1}^{\operatorname{dim}V}b_{j}v_{j},\sum_{i=1}^{\operatorname{dim}V}a_{i}v_{i}\right\rangle}.$$
2. (ii)

   	$\displaystyle\left\langle\alpha\sum_{i=1}^{\operatorname{dim}V}a_{i}v_{i}+\beta\sum_{j=1}^{\operatorname{dim}V}b_{j}v_{j},\sum_{k=1}^{\operatorname{dim}V}c_{k}v_{k}\right\rangle=\left\langle\sum_{i=1}^{\operatorname{dim}V}(\alpha a_{i}+\beta b_{i})v_{i},\sum_{k=1}^{\operatorname{dim}V}c_{k}v_{k}\right\rangle=\sum_{i=1}^{\operatorname{dim}V}(\alpha a_{i}+\beta b_{i})\overline{c_{i}}$
   	$\displaystyle=\alpha\sum_{i=1}^{\operatorname{dim}V}a_{i}\overline{c_{i}}+\beta\sum_{i=1}^{\operatorname{dim}V}b_{i}\overline{c_{i}}=\alpha\left\langle\alpha\sum_{i=1}^{\operatorname{dim}V}a_{i}v_{i},\sum_{k=1}^{\operatorname{dim}V}c_{k}v_{k}\right\rangle+\beta\left\langle+\beta\sum_{j=1}^{\operatorname{dim}V}b_{j}v_{j},\sum_{k=1}^{\operatorname{dim}V}c_{k}v_{k}\right\rangle.$

3. (iii)
   $$\left\langle\sum_{i=1}^{\operatorname{dim}V}a_{i}v_{i},\sum_{j=1}^{\operatorname{dim}V}a_{j}v_{j}\right\rangle=\sum_{i=1}^{\operatorname{dim}V}a_{i}\overline{a_{i}}=\sum_{i=1}^{\operatorname{dim}V}|a_{i}|^{2}\geq 0,$$

   ​

   with equality if and only if all $a_{i}$ are zero, i.e. $\sum_{i=1}^{\operatorname{dim}V}a_{i}v_{i}=0$.

Solution: We verify that $\langle\cdot,\cdot\rangle_{G}$ is an inner product:

1. (i)
   $$\langle v,w\rangle_{G}=\frac{1}{|G|}\sum_{g\in G}\langle\pi(g)v,\pi(g)w\rangle=\frac{1}{|G|}\sum_{g\in G}\overline{\langle\pi(g)w,\pi(g)v\rangle}=\overline{\frac{1}{|G|}\sum_{g\in G}\langle\pi(g)w,\pi(g)v\rangle}=\overline{\langle w,v\rangle_{G}}.$$
2. (ii)

   	$\displaystyle\left\langle\alpha u+\beta v,w\right\rangle_{G}=\frac{1}{|G|}\sum_{g\in G}\langle\pi(g)(\alpha u+\beta v),\pi(g)w\rangle=\frac{1}{|G|}\sum_{g\in G}\alpha\langle\pi(g)u,\pi(g)w\rangle+\beta\langle\pi(g)v,\pi(g)w\rangle$
   	$\displaystyle=\alpha\frac{1}{|G|}\sum_{g\in G}\langle\pi(g)u,\pi(g)w\rangle+\beta\frac{1}{|G|}\sum_{g\in G}\langle\pi(g)v,\pi(g)w\rangle=\alpha\langle u,w\rangle+\beta\langle v,w\rangle.$

3. (iii)
   $$\langle v,v\rangle_{G}=\frac{1}{|G|}\sum_{g\in G}\langle\pi(g)v,\pi(g)v\rangle.$$

   ​

   For each $g\in G$, we have $\langle\pi(g)v,\pi(g)\rangle\geq 0$ (since $\langle\cdot,\cdot\rangle$ is an inner product). Each $\pi(g)$ is invertible, so $\pi(g)v=0$ if and only if $v=0$, hence

   $$\frac{1}{|G|}\sum_{g\in G}\langle\pi(g)v,\pi(g)v\rangle\geq 0,$$

   as the sum of a collection of non-negative numbers is non-negative, and equality is obtained if and only if $v=0$.

Letting $(\pi,V)$, $G$, and $\langle\cdot,\cdot\rangle_{G}$ be as in Problem 28, show

Solution:

Recalling that $V=W\oplus W^{\perp}$, this shows that $(\pi,V)=(\pi,W)\oplus(\pi,W^{\perp})$, proving Theorem 10!

Solution: Given any $u\in W^{\perp}$ and $g\in G$, we need to show that $\pi(g)u\in W^{\perp}$. For any $w\in W$, we have

Since $W$ is a subrepresentation, $\pi(g^{-1})w\in W$. By assumption, $u\in W^{\perp}$, hence $\langle\pi(g^{-1})w,u\rangle_{G}=0$, giving

hence $\pi(g)u\in W^{\perp}$.

Remark: a representation of $(\pi,V)$ of a group $G$ on an inner product space $V$ (with inner product $\langle\cdot,\cdot\rangle$) is called unitary if $\langle\pi(g)v,\pi(g)w\rangle=\langle v,w\rangle$ for all $g\in G,\,v,w\in V$. Problem 28 shows that we may always assume that a finite-dimensional representation of a finite group is unitary.

Solution: Since the ${\bf b}_{i}$ form an ON-basis, the matrix of $\pi(g^{-1})$ with respect to this basis is given by the formula

(here $i$ runs through the rows, and $j$ the columns). The conjugate transpose of this matrix is then given by the formula

which is precisely the formula for $[\pi(g)]_{{\mathcal{B}}}$. We have thus shown that

Taking the conjugate transpose of both sides, together with the identity $(M^{*})^{*}=M$ gives that $\pi(g)$ is unitary.

This problem gives an alternative way to show that a finite-order matrix $M\in\operatorname{GL}_{n}({\mathbb{C}})$ (say, of order $d$) is diagonalisable: consider the finite group $\langle M\rangle=\{I,M,M^{2},\ldots,M^{d-1}\}$. Form a new invariant inner product $\langle\cdot,\cdot\rangle_{M}$ on ${\mathbb{C}}^{n}$ by

As seen above, choosing an orthonormal basis ${\mathcal{B}}$ of ${\mathbb{C}}^{n}$ with respect to $\langle\cdot,\cdot\rangle_{M}$, and using the change of basis formula to compute the matrix of $v\mapsto Mv$ with respect to this basis gives a unitary matrix $[M]_{{\mathcal{B}}}$. The spectral theorem applies to unitary matrices, allowing us to write $[M]_{{\mathcal{B}}}=B\operatorname{diag}(\lambda_{1},\ldots,\lambda_{n})B^{-1},$ and then $M=[{\mathcal{B}}]B\operatorname{diag}(\lambda_{1},\ldots,\lambda_{n})B^{-1}[{\mathcal{B}}]^{-1}$.

Use Theorem 16 (ii) to verify that the table contains no redundancies, and that every entry must in fact define a representation of $D_{n}$.

Solution: By Theorem 16 (ii), $\sum_{\sigma\in\mathrm{Irr}(D_{n})}(\operatorname{dim}\sigma)^{2}=|D_{n}|=2n$. Summing the dimensions squared of the entries of the table gives $1^{2}+1^{2}+2^{2}\cdot\frac{(n-1)}{2}=2n.$ Since we proved in class that any irreducible representation of $D_{n}$ must have $\pi(r)$ and $\pi(s)$ as given by the table. If any of the entries of the table were isomorphic, or were not legitimate representations, we would be able to remove these from the list to obtain a new list that had precisely one representative from each isomorphism class of each irreducible representation. However, the sums of squares of dimensions of this new list could not possibly be equal to $2n$, as it was obtained by removing entries from our current list. Thus it is not possible to remove any entries of our table and still have a representative of each isomorphism class.

Solution: We first verify that $T^{\prime}$ is a linear operator: for $\alpha,\beta\in{\mathbb{C}}$ and $v_{1},v_{2}\in V$,

hence $T^{\prime}\in\operatorname{Hom}(V,U)$. We next observe that for $S,T\in\operatorname{Hom}(W,U)$ and $\alpha,\beta\in{\mathbb{C}}$, and $v\in V$ we have

hence $T\mapsto T^{\prime}$ is a linear map from $\operatorname{Hom}(W,U)$ to $\operatorname{Hom}(V,U)$. We now verify that this map sends $\operatorname{Hom}_{G}(W,U)$ to $\operatorname{Hom}_{H}(V,U)$: let $T\in\operatorname{Hom}_{G}(W,U)$. Then for any $h\in H$ and $v\in V$, we have

i.e. $T^{\prime}$ is an $H$-intertwiner, as required.

Solution: We have

Using Problem 79, we may rewrite this as follows:

This shows that for each polytabloid $t$ and permutation $\sigma$, $\pi(\sigma)\mathbf{e}_{t}=\mathbf{e}_{\sigma\cdot t}\in{\mathcal{S}}^{\lambda}$. The vectors $\mathbf{e}_{t}$ span ${\mathcal{S}}^{\lambda}$, hence ${\mathcal{S}}^{\lambda}$ is closed under the action of $S_{n}$, and is therefore a subrepresentation.